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A uniform circular hoop rotates about a peg on a perfectly smooth horizontal plane; find the angular velocity of the hoop if the peg is suddenly removed and simultaneously another peg is introduced, about which it begins to rotate.
Let 0 and O1', Fig. 125, be the positions of the first and second peg, respectively. The circle in continuous line may be considered to represent the position of the hoop just ^_____
before it stops rotating about 0 and just after it begins to rotate about 0'.
The only force which comes into play when the hoop strikes the peg 0' passes through 0', hence it produces no effect upon the angular momentum about 0;. Therefore the angular momentum about 0' just after the hoop strikes the peg equals the angular momentum just before. The angular momentum after the hoop begins to rotate about 0' is
H'0> = /«' = 2 ma V,
where Hr0' is the angular momentum and «' the angular velocity about the 0', m the mass, and a the radius of the hoop.
The angular momentum about 0' just before the hoop begins to rotate about 0' equals the angular momentum of the hoop due to the motion of the hoop about its geometrical axis plus its angular momentum due to its motion with its center of mass. Therefore HOr = Ic co + mv • a cos a
— ma2co + ma2co cos a
— wa2co (1 + cos a),
where co is the angular velocity about the peg 0, and a the angle which the arc 00' subtends at the center of the hoop. But since
ff V = Ho', 2 ma2co' = ma2co (1 + cos a)
i , 1 + cos a and. co = —•--------co